It was shown in [1]} that there exists \(\mu ^{\prime }\in \mathcal {E}^+\)  uniquely determined by any of (a)–(d) (with \(\mu ^A\)  replaced by \(\mu ^{\prime }\) ), and that it satisfies relations (REF )–() (again with \(\mu ^A\)  replaced by \(\mu ^{\prime }\) ).A kernel \(\kappa \)  was defined in [1]} to be positive; however, this restriction on \(\kappa \)  is unnecessary for the validity of the above-quoted results from [1]}.
It thus remains to show that the same \(\mu ^{\prime }\)  meets Definition REF , and so it serves as the (unique) inner balayage \(\mu ^A\) . To this end, one can assume the two conditions
\(c_*(A)>0\text{ \ and \ }\kappa \mu \ne 0\text{ \ n.e.\ on $A$}\) 
to be fulfilled, for if not, zero measure belongs to \(\Gamma _{A,\mu }^+\) , and therefore
\(\mu ^A=0=\mu ^{\prime },\) 
the former equality being obvious from Definition REF , while the latter being clear from (a).
Applying (c), we now conclude from (REF ) that then necessarily
\(\mu ^{\prime }\ne 0.\) 
As \(\mu ^{\prime }\in \Gamma _{A,\mu }^+\)  by (REF ), the theorem will follow once we show that
\(\Vert \mu ^{\prime }\Vert \leqslant \Vert \nu \Vert ,\) 
\(\nu \in \Gamma _{A,\mu }^+\)  being arbitrarily given. But, by virtue of (a), \(\kappa \mu ^{\prime }\leqslant \kappa \nu \)  everywhere on \(X\) ,
which yields, by the Cauchy–Schwarz (Bunyakovski) inequality,
\(\Vert \mu ^{\prime }\Vert ^2=\int \kappa \mu ^{\prime }\,d\mu ^{\prime }\leqslant \int \kappa \nu \,d\mu ^{\prime }=\langle \nu ,\mu ^{\prime }\rangle \leqslant \Vert \nu \Vert \cdot \Vert \mu ^{\prime }\Vert ,\) 
and dividing by \(\Vert \mu ^{\prime }\Vert \ne 0\)  (cf. (REF )) results in (REF ).
